Solve for $x$ : $5x^2 + 20x - 160 = 0$
Explanation: Dividing both sides by $5$ gives: $ x^2 + {4}x {-32} = 0 $ The coefficient on the $x$ term is $4$ and the constant term is $-32$ , so we need to find two numbers that add up to $4$ and multiply to $-32$ The two numbers $-4$ and $8$ satisfy both conditions: $ {-4} + {8} = {4} $ $ {-4} \times {8} = {-32} $ $(x {-4}) (x + {8}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -4) (x + 8) = 0$ $x - 4 = 0$ or $x + 8 = 0$ Thus, $x = 4$ and $x = -8$ are the solutions.